发布时间 : 星期六 文章不确定性推理部分参考答案更新完毕开始阅读24174449a22d7375a417866fb84ae45c3b35c2e7
第6章 不确定性推理部分参考答案
6.8 设有如下一组推理规则: r1: IF E1 THEN E2 (0.6)
r2: IF E2 AND E3 THEN E4 (0.7) r3: IF E4 THEN H (0.8) r4: IF E5 THEN H (0.9)
且已知CF(E1)=0.5, CF(E3)=0.6, CF(E5)=0.7。求CF(H)=? 解：(1) 先由r1求CF(E2)
CF(E2)=0.6 × max{0,CF(E1)} =0.6 × max{0,0.5}=0.3
(2) 再由r2求CF(E4)
CF(E4)=0.7 × max{0, min{CF(E2 ), CF(E3 )}} =0.7 × max{0, min{0.3, 0.6}}=0.21
(3) 再由r3求CF1(H)
CF1(H)= 0.8 × max{0,CF(E4)}
=0.8 × max{0, 0.21)}=0.168 (4) 再由r4求CF2(H)
CF2(H)= 0.9 ×max{0,CF(E5)} =0.9 ×max{0, 0.7)}=0.63
(5) 最后对CF1(H )和CF2(H)进行合成，求出CF(H) CF(H)= CF1(H)+CF2(H)+ CF1(H) × CF2(H) =0.692
6.10 设有如下推理规则
r1: IF E1 THEN (2, 0.00001) H1 r2: IF E2 THEN (100, 0.0001) H1 r3: IF E3 THEN (200, 0.001) H2 r4: IF H1 THEN (50, 0.1) H2
且已知P(E1)= P(E2)= P(H3)=0.6, P(H1)=0.091, P(H2)=0.01, 又由用户告知： P(E1| S1)=0.84, P(E2|S2)=0.68, P(E3|S3)=0.36 请用主观Bayes方法求P(H2|S1, S2, S3)=? 解：(1) 由r1计算O(H1| S1)
先把H1的先验概率更新为在E1下的后验概率P(H1| E1) P(H1| E1)=(LS1 × P(H1)) / ((LS1-1) × P(H1)+1) =(2 × 0.091) / ((2 -1) × 0.091 +1) =0.16682
由于P(E1|S1)=0.84 > P(E1)，使用P(H | S)公式的后半部分，得到在当前观察S1下的后验概率P(H1| S1)和后验几率O(H1| S1)
P(H1| S1) = P(H1) + ((P(H1| E1) – P(H1)) / (1 - P(E1))) × (P(E1| S1) – P(E1)) = 0.091 + (0.16682 –0.091) / (1 – 0.6)) × (0.84 – 0.6) =0.091 + 0.18955 × 0.24 = 0.136492 O(H1| S1) = P(H1| S1) / (1 - P(H1| S1)) = 0.15807 (2) 由r2计算O(H1| S2)
先把H1的先验概率更新为在E2下的后验概率P(H1| E2) P(H1| E2)=(LS2 × P(H1)) / ((LS2-1) × P(H1)+1) =(100 × 0.091) / ((100 -1) × 0.091 +1) =0.90918
由于P(E2|S2)=0.68 > P(E2)，使用P(H | S)公式的后半部分，得到在当前观察S2下的后验概率P(H1| S2)和后验几率O(H1| S2)
P(H1| S2) = P(H1) + ((P(H1| E2) – P(H1)) / (1 - P(E2))) × (P(E2| S2) – P(E2)) = 0.091 + (0.90918 –0.091) / (1 – 0.6)) × (0.68 – 0.6) =0.25464
O(H1| S2) = P(H1| S2) / (1 - P(H1| S2)) =0.34163
(3) 计算O(H1| S1,S2)和P(H1| S1,S2) 先将H1的先验概率转换为先验几率
O(H1) = P(H1) / (1 - P(H1)) = 0.091/(1-0.091)=0.10011
再根据合成公式计算H1的后验几率
O(H1| S1,S2)= (O(H1| S1) / O(H1)) × (O(H1| S2) / O(H1)) × O(H1) = (0.15807 / 0.10011) × (0.34163) / 0.10011) × 0.10011 = 0.53942
再将该后验几率转换为后验概率
P(H1| S1,S2) = O(H1| S1,S2) / (1+ O(H1| S1,S2)) = 0.35040 (4) 由r3计算O(H2| S3)
先把H2的先验概率更新为在E3下的后验概率P(H2| E3) P(H2| E3)=(LS3 × P(H2)) / ((LS3-1) × P(H2)+1) =(200 × 0.01) / ((200 -1) × 0.01 +1) =0.09569
由于P(E3|S3)=0.36 < P(E3)，使用P(H | S)公式的前半部分，得到在当前观察S3下的后验概率P(H2| S3)和后验几率O(H2| S3)
P(H2| S3) = P(H2 | ? E3) + (P(H2) – P(H2| ?E3)) / P(E3)) × P(E3| S3) 由当E3肯定不存在时有
P(H2 | ? E3) = LN3 × P(H2) / ((LN3-1) × P(H2) +1) = 0.001 × 0.01 / ((0.001 - 1) × 0.01 + 1) = 0.00001 因此有
P(H2| S3) = P(H2 | ? E3) + (P(H2) – P(H2| ?E3)) / P(E3)) × P(E3| S3) =0.00001+((0.01-0.00001) / 0.6) × 0.36 =0.00600
O(H2| S3) = P(H2| S3) / (1 - P(H2| S3))
=0.00604
(5) 由r4计算O(H2| H1)
先把H2的先验概率更新为在H1下的后验概率P(H2| H1) P(H2| H1)=(LS4 × P(H2)) / ((LS4-1) × P(H2)+1) =(50 × 0.01) / ((50 -1) × 0.01 +1) =0.33557
由于P(H1| S1,S2)=0.35040 > P(H1)，使用P(H | S)公式的后半部分，得到在当前观察S1,S2下H2的后验概率P(H2| S1,S2)和后验几率O(H2| S1,S2)
P(H2| S1,S2) = P(H2) + ((P(H2| H1) – P(H2)) / (1 - P(H1))) × (P(H1| S1,S2) – P(H1)) = 0.01 + (0.33557 –0.01) / (1 – 0.091)) × (0.35040 – 0.091) =0.10291
O(H2| S1,S2) = P(H2| S1, S2) / (1 - P(H2| S1, S2)) =0.10291/ (1 - 0.10291) = 0.11472 (6) 计算O(H2| S1,S2,S3)和P(H2| S1,S2,S3) 先将H2的先验概率转换为先验几率
O(H2) = P(H2) / (1 - P(H2) )= 0.01 / (1-0.01)=0.01010
再根据合成公式计算H1的后验几率
O(H2| S1,S2,S3)= (O(H2| S1,S2) / O(H2)) × (O(H2| S3) / O(H2)) ×O(H2) = (0.11472 / 0.01010) × (0.00604) / 0.01010) × 0.01010 =0.06832
再将该后验几率转换为后验概率
P(H2| S1,S2,S3) = O(H1| S1,S2,S3) / (1+ O(H1| S1,S2,S3)) = 0.06832 / (1+ 0.06832) = 0.06395
可见，H2原来的概率是0.01，经过上述推理后得到的后验概率是0.06395，它相当于先验概率的6倍多。
6.11设有如下推理规则
r1: IF E1 THEN (100, 0.1) H1 r2: IF E2 THEN (50, 0.5) H2 r3: IF E3 THEN (5, 0.05) H3
且已知P(H1)=0.02, P(H2)=0.2, P(H3)=0.4，请计算当证据E1，E2，E3存在或不存在时P(Hi | Ei)或P(Hi |﹁Ei)的值各是多少(i=1, 2, 3)？
解：(1) 当E1、E2、E3肯定存在时，根据r1、r2、r3有
P(H1 | E1) = (LS1 × P(H1)) / ((LS1-1) × P(H1)+1)
= (100 × 0.02) / ((100 -1) × 0.02 +1) =0.671
P(H2 | E2) = (LS2 × P(H2)) / ((LS2-1) × P(H2)+1)
= (50 × 0.2) / ((50 -1) × 0.2 +1)
=0.9921
P(H3 | E3) = (LS3 × P(H3)) / ((LS3-1) × P(H3)+1)
= (5 × 0.4) / ((5 -1) × 0.4 +1)
=0.769
(2) 当E1、E2、E3肯定存在时，根据r1、r2、r3有
P(H1 | ?E1) = (LN1 × P(H1)) / ((LN1-1) × P(H1)+1)
= (0.1 × 0.02) / ((0.1 -1) × 0.02 +1) =0.002
P(H2 | ?E2) = (LN2 × P(H2)) / ((LN2-1) × P(H2)+1)
= (0.5 × 0.2) / ((0.5 -1) × 0.2 +1) =0.111
P(H3 | ?E3) = (LN3 × P(H3)) / ((LN3-1) × P(H3)+1)
= (0.05 × 0.4) / ((0.05 -1) × 0.4 +1) =0.032
6.13 设有如下一组推理规则:
r1: IF E1 AND E2 THEN A={a} (CF={0.9})
r2: IF E2 AND (E3 OR E4) THEN B={b1, b2} (CF={0.8, 0.7}) r3: IF A THEN H={h1, h2, h3} (CF={0.6, 0.5, 0.4}) r4: IF B THEN H={h1, h2, h3} (CF={0.3, 0.2, 0.1}) 且已知初始证据的确定性分别为：
CER(E1)=0.6, CER(E2)=0.7, CER(E3)=0.8, CER(E4)=0.9。
假设|Ω|=10，求CER(H)。 解：其推理过程参考例6.9 具体过程略
6.15 设
U=V={1，2，3，4}
且有如下推理规则：
IF x is 少 THEN y is 多 其中，“少”与“多”分别是U与V上的模糊集，设 少=0.9/1+0.7/2+0.4/3 多=0.3/2+0.7/3+0.9/4 已知事实为
x is 较少 “较少”的模糊集为
较少=0.8/1+0.5/2+0.2/3 请用模糊关系Rm求出模糊结论。